Derive the expression for the self inductance of a long solenoid of cr

1.	Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.
Answer» Magnetic field inside the solenoid B =μ₀nI n → number of turns per unit length, I → current flowing, ϕ' = Magnetic field through each turn = BA = μ₀nIA Total magnetic flux linked with solenoid ϕ = Nϕ' = nl x μ₀nIA =μ₀n²AlI LI =>L =μ₀n²Al L → self inductance of solenoid and l → length of solenoid.

Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.

Answer»

Magnetic field inside the solenoid

B =μ₀nI
n → number of turns per unit length, I → current flowing,
ϕ' = Magnetic field through each turn = BA = μ₀nIA
Total magnetic flux linked with solenoid
ϕ = Nϕ' = nl x μ₀nIA =μ₀n²AlI LI =>L =μ₀n²Al
L → self inductance of solenoid and l → length of solenoid.

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Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.

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